By \(u(u_0) = u(u_0)(t)\)  we denote the solution to
\(\partial _t^{\alpha } u(t) = Au(t), \quad u(0)=u_0.\) 
Then we know that for \(u_0\in H\) , there exists a unique solution
\(u(u_0) \in C([0,T];H)\)  (e.g., Sakamoto and Yamamoto [1]} for
\(0<\alpha <1\)  and Pazy [2]} for \(\alpha =1\) ).
Also, it is known that
\(\Vert u(u_0)(t) - u_0\Vert ^2 ={\left\lbrace \begin{array}{ll}\displaystyle \sum _{n=1}^{\infty } \langle u_0, \varphi _n \rangle ^2(E_{\alpha ,1}(-\lambda _nt^{\alpha }) -1)^2, & 0<\alpha <1,\\\displaystyle \sum _{n=1}^{\infty } \langle u_0, \varphi _n \rangle ^2(\mathrm {e}^{-\lambda _nt} -1)^2, & \alpha =1.\end{array}\right.}\) 
Assume that the conclusion of the theorem does not hold.
By \(\theta (u_0)\) , we denote the exponent \(\theta \)  in (5).
Then there exist \(u_0^k \in M\)  such that \(\lim _{k\rightarrow \infty }\theta (u_0^k) = 0\) .
Set \(\xi _k:= 2T\theta (u_0^k)\) .
Moreover, the compactness implies that a subsequence,
still denoted by \(u_0^k\) , satisfies \(\lim _{k\rightarrow \infty }u_0^k= u_0 \in H\)  with some \(u_0\) .
In
\(\int ^T_0 (T-s)^{\alpha -1}\Vert u_k(s)\Vert \, \mathrm {d}s =\Vert u_k(\xi _k)\Vert \frac{T^{\alpha }}{\alpha },\) 
we set \(u_k(t):= u(u_0^k)(t)\)  for \(k\in \mathbb {N}\)  and \(t>0\) .
We write \(u=u(u_0)\) .
We discuss only the case \(0<\alpha <1\) , and the case \(\alpha =1\)  can be
discussed in the same way.
We have
\(& \Vert u_k(\xi _k) - u(0)\Vert = \Vert u_k(\xi _k) - u_k(0) + u_k(0) - u(0)\Vert \\\le &\Vert u_k(\xi _k) - u_k(0)\Vert + \Vert u_k(0) - u_0\Vert ,\) 

and
\(\Vert u_k(\xi _k) - u_k(0)\Vert ^2= \sum _{n=1}^{\infty } \langle u_0^k, \varphi _n \rangle ^2(E_{\alpha ,1}(-\lambda _n\xi _k^{\alpha }) -1)^2\) 
and
\(\lim _{k\rightarrow \infty } \Vert u_k(0) - u_0\Vert = \lim _{k\rightarrow \infty } \Vert u_0^k - u_0\Vert = 0.\) 
Since \(\Vert Au_0^k\Vert \le R\)  for all \(k\in \mathbb {N}\) , we have
\(& \vert \langle u_0^k, \varphi _n \rangle \vert = \vert \langle A^{-1}Au_0^k, \varphi _n\rangle \vert = \vert \langle Au_0^k, A^{-1}\varphi _n\rangle \vert = \left|\frac{1}{\lambda _n}\langle Au_0^k, \varphi _n\rangle \right|\\\le & \frac{1}{\lambda _n}\Vert Au_0^k\Vert \Vert \varphi _n\Vert \le \frac{R}{\lambda _n}.\) 

Also using the assumption \(\sum _{n=1}^\infty \frac{1}{\lambda _n^2} < \infty \) , we
can
apply the Lebesgue convergence theorem to see that
\(\lim _{k\rightarrow \infty } \Vert u_k(\xi _k) - u_k(0)\Vert = 0\) .
Therefore, letting \(k \rightarrow \infty \)  in (REF ),
we obtain
\(\int ^T_0 (T-s)^{\alpha -1}(\Vert u(s)\Vert -\Vert u(0)\Vert )\, \mathrm {d}s = 0.\) 
By \(\lambda _n> 0\) , using
\(\Vert u(s)\Vert ^2 = \sum _{n=1}^{\infty } \vert \langle u_0, \varphi _n\rangle \vert ^2 E_{\alpha ,1}(-\lambda _ns^{\alpha })^2\le \sum _{n=1}^{\infty } \vert \langle u_0, \varphi _n \rangle \vert ^2,\) 
we see that \(\Vert u(s)\Vert \le \Vert u(0)\Vert \)  for
\(s>0\) . Hence \(\Vert u(s)\Vert = \Vert u(0)\Vert \)  for all \(s>0\) .
Therefore the eigenvectors expansion yields
\(\sum _{n=1}^{\infty } \vert \langle u_0, \varphi _n\rangle \vert ^2(1 - E_{\alpha ,1}(-\lambda _ns^{\alpha }))^2 = 0.\) 
Since \(\lambda _n > 0\) , we have \(E_{\alpha ,1}(-\lambda _ns^{\alpha }) < 1\)  for all
\(n\in \mathbb {N}\) . Therefore, \(\langle u_0, \varphi _n\rangle = 0\)  for all
\(n\in \mathbb {N}\) , which means that \(u_0 = 0\) .
Since \(\lim _{k\rightarrow \infty } \Vert u_0^k\Vert = \Vert u_0\Vert \)  and
\(\Vert u_0^k\Vert \ge \delta \)  for \(k\in \mathbb {N}\)  by \(u_0^k \in M\) , we see that \(0 \ge \delta \) . Hence we reach a contradiction. Thus the proof of Proposition REF  is complete.
